In a Rubik's cube what is the probability of randomly getting a face blue? When we solve a Rubik's cube we make planned moves and easily get the desired results but here I am talking about getting a colour  by randomly moving a Rubik's cube without making an effort to solve it.
I am getting 1/269,438,400 as the probability but I am not sure that it's correct.
I solved the problem this way. I started with the middle block and considered the probability of it being blue as 1/6. Then I considered the blocks located b/w 2 corner blocks. For the first such block the probability is 4/24 that is 1/6 as 4 such blocks are available and a total of 24 blocks of that position are available but for the next such block I considered the probability as 3/22 as in the previous step 2 of the total such blocks because unavailable and one desired (blue) block of that position became unavailable. This way I got the probabilities as 1/10 and 1/18.
I did the same for corner blocks. I then multiplied all the values I got as they are all interdependent.
 A: This is a rather involved problem which I rather doubt you would have correctly taken into account everything.  It is not clear where your calculation comes from.
As for an approach... you can imagine the center pieces never move and use those as a central focal point.  Now... for the side with the blue center square to have all of the blue stickers on that face, two things must happen.  The four corners must all be the corners who have a blue side.  Then, they must all be oriented such that they have the blue side on the correct side.  If we were to assume each collection of four corners is equally likely to be chosen and each orientation of each is equally likely to be used this occurs with probability $\frac{1}{\binom{8}{4}}\times\frac{1}{3^4}$.
Next, the four edges must all be the edges who have a blue edge and they must all be oriented the correct way.  If we were to again assume each collection of four edges are equally likely to be chosen and each orientation of each is equally likely this would be occurring with probability $\frac{1}{\binom{12}{4}}\times \frac{1}{2^4}$.
I get as an estimation then:
$$\frac{1}{\binom{8}{4}\binom{12}{4}6^4}\approx \frac{1}{44906400}\approx 2.23\times 10^{-8}$$
rather far from your $\approx 3.7\times 10^{-9}$
Now, to be fair, we know that there must be some dependency between the events if we were to have extended it to all of the positions elsewhere on the cube including on the sides not touching the intended blue side since we know that there are arrangements which cannot be achieved via standard moves, however I believe if looking solely at the blue face they may yet be independent.  Ensuring that we remained within the orbit can be done by enforcing the orientation of one of the opposite corners and opposite edges.
A: Your problem is unclear, but from your attempt it seems you allow slice turns, and are looking at a fixed face, so that indeed the probability that the centre piece is blue is $1/6$. I have no clear idea what you are trying to do after that, but simply consider that all permutations of the edge pieces are equally likely given fixed centres, so the probability that the 4 relevant edge pieces have blue is $1/\binom{12}{4}$. The probability that those 4 edge pieces have blue facing the right way is $1/2^4$, since all flip combinations are equally likely (note that this does not hold if you want to look at all edge pieces; it holds here only because we are looking at only $4$ of them). This does give the same answer $1/495$ as you obtained, so I suppose you must have reasoned correctly for that. For the corner pieces, similarly probability of having blue is $1/\binom{8}{4}$ and probability of their blue facing the right way is $1/3^4$. Again, it is not trivial that the edge-pieces-blue and corner-pieces-blue events are independent, but is true because the parity restriction has no effect when we are only looking at one face-full of pieces. Final answer is indeed $1/269438400$ as you stated.
